Lời giải:
$1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x(x+1)}=\frac{2014}{2015}$
$\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x(x+1)}=\frac{2014}{2015}$
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1007}{2015}$
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1007}{2015}$
$1-\frac{1}{x+1}=\frac{1007}{2015}$
$\frac{1}{x+1}=1-\frac{1007}{2015}=\frac{1008}{2015}$
$\Rightarrow x+1=\frac{2015}{1008}$
$\Rightarrow x=\frac{1007}{1008}$
\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x(x+1)}=\frac{2014}{2015}$
